7.3: Exercises

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Page ID: 48991

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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Exercise $\PageIndex{1}$

Use the horizontal line test to determine if the function is one-to-one.

$clipboard_e00dcca2039e18db8fefb626195322c1a.png$
$clipboard_e41a01f3ceb2fa0e71d04b9a4c5220ae3.png$
$f(x)=x^2+2x+5$
$f(x)=x^2-14 x+29$
$f(x)=x^3-5x^2$
$f(x)=\dfrac{x^2}{x^2-3}$
$f(x)=\sqrt{x+2}$
$f(x)=\sqrt{|x+2|}$

Answer

no (that is: the function is not one-to-one)
yes
no
no
no
no
yes
no

Exercise $\PageIndex{2}$

Find the inverse of the function $f$ and check your solution.

$f(x)=4x+9$
$f(x)=-8x-3$
$f(x)=\sqrt{x+8}$
$f(x)=\sqrt{3x+7}$
$f(x)=6\cdot \sqrt{-x-2}$
$f(x)=x^3$
$f(x)=(2x+5)^3$
$f(x)=2\cdot x^3+5$
$f(x)=\dfrac{1}{x}$
$f(x)=\dfrac{1}{x-1}$
$f(x)=\dfrac{1}{\sqrt{x-2}}$
$f(x)=\dfrac{-5}{4-x}$
$f(x)=\dfrac{x}{x+2}$
$f(x)=\dfrac{3x}{x-6}$
$f(x)=\dfrac{x+2}{x+3}$
$f(x)=\dfrac{7-x}{x-5}$
$f$ given by the table: $\begin{array}{|c||c|c|c|c|c|c|}
\hline x & 2 & 4 & 6 & 8 & 10 & 12 \\
\hline \hline f(x) & 3 & 7 & 1 & 8 & 5 & 2 \\
\hline
\end{array} $

Answer

$f^{-1}(x)=\dfrac{x-9}{4}$
$f^{-1}(x)=-\dfrac{x+3}{8}$
$f^{-1}(x)=x^{2}-8$
$f^{-1}(x)=\dfrac{x^{2}-7}{3}$
$f^{-1}(x)=-\left(\dfrac{x}{6}\right)^{2}-2=\dfrac{-x^{2}-72}{36}$
$f^{-1}(x)=\sqrt[3]{x}$
$f^{-1}(x)=\dfrac{\sqrt[3]{x}-5}{2}$
$f^{-1}(x)=\sqrt[3]{\dfrac{x-5}{2}}$
$f^{-1}(x)=\dfrac{1}{x}+1=\dfrac{1+x}{x}$
$f^{-1}(x)=\left(\dfrac{1}{x}\right)^{2}+2=\dfrac{1+2 x^{2}}{x^{2}}$
$f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}$
$f^{-1}(x)=\dfrac{5}{y}+4=\dfrac{5+4 y}{y}$
$f^{-1}(x)=\dfrac{2 x}{1-x}$
$f^{-1}(x)=\dfrac{6 x}{x-3}$
$f^{-1}(x)=\dfrac{2-3 x}{x-1}$
$f^{-1}(x)=\dfrac{5 x+7}{x+1}$
$\begin{array}{|c||c|c|c|c|c|c|}
\hline x & 3 & 7 & 1 & 8 & 5 & 2 \\
\hline \hline f^{-1}(x) & 2 & 4 & 6 & 8 & 10 & 12 \\
\hline
\end{array}$

Exercise $\PageIndex{3}$

Restrict the domain of the function $f$ in such a way that $f$ becomes a one-to-one function. Find the inverse of $f$ with the restricted domain.

$f(x)=x^2$
$f(x)=(x+5)^2+1$
$f(x)=|x|$
$f(x)=|x-4|-2$
$f(x)=\dfrac{1}{x^2}$
$f(x)=\dfrac{-3}{(x+7)^2}$
$f(x)=x^4$
$f(x)=\dfrac{(x-3)^4}{10}$

Answer

restricting to the domain $D=[0, \infty)$ gives the inverse $f^{-1}(x)=\sqrt{x}$
restricting to the domain $D=[-5, \infty)$ gives the inverse $f^{-1}(x)=\sqrt{x-1}-5$
restricting to the domain $D=[0, \infty)$ gives the inverse $f^{-1}(x)=x$
restricting to the domain $D=[4, \infty)$ gives the inverse $f^{-1}(x)=x+6$
restricting to the domain $D=[0, \infty)$ gives the inverse $f^{-1}(x)=\sqrt{\dfrac{1}{x}}$
restricting to the domain $D=[-7, \infty)$ gives the inverse $f^{-1}(x)=\sqrt{-\dfrac{3}{x}}-7$
restricting to the domain $D=[0, \infty)$ gives the inverse $f^{-1}(x)=\sqrt[4]{x}$
restricting to the domain $D=[3, \infty)$ gives the inverse $f^{-1}(x)=3+\sqrt[4]{10 x}$